Simplify the following expression: $y = \dfrac{-7x^2- 5x+2}{-7x + 2}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-7)}{(2)} &=& -14 \\ {a} + {b} &=& &=& {-5} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-14$ and add them together. Remember, since $-14$ is negative, one of the factors must be negative. The factors that add up to ${-5}$ will be your ${a}$ and ${b}$ When ${a}$ is ${2}$ and ${b}$ is ${-7}$ $ \begin{eqnarray} {ab} &=& ({2})({-7}) &=& -14 \\ {a} + {b} &=& {2} + {-7} &=& -5 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-7}x^2 +{2}x) + ({-7}x +{2}) $ Factor out the common factors: $ x(-7x + 2) + 1(-7x + 2)$ Now factor out $(-7x + 2)$ $ (-7x + 2)(x + 1)$ The original expression can therefore be written: $ \dfrac{(-7x + 2)(x + 1)}{-7x + 2}$ We are dividing by $-7x + 2$ , so $-7x + 2 \neq 0$ Therefore, $x \neq \frac{2}{7}$ This leaves us with $x + 1; x \neq \frac{2}{7}$.